Expert Integration Help: Simplifying \frac{1}{(1+sinx)} in Minutes

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Homework Help Overview

The discussion revolves around integrating the expression \(\frac{1}{(1+\sin x)}\), with participants exploring various methods and substitutions related to trigonometric functions.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss different integration techniques, including substitution methods and multiplying by conjugates. Some express uncertainty about their approaches, while others suggest alternative methods to simplify the problem.

Discussion Status

Several participants have offered guidance on potential methods, such as multiplying the numerator and denominator by \((1-\sin x)\) or using the substitution \(u = \tan(x/2)\). There is a mix of interpretations and attempts, with some participants reporting success with the suggested approaches.

Contextual Notes

Some participants indicate that they are new to integration, which may affect their confidence in the methods being discussed. There is also mention of specific trigonometric identities and their application in the integration process.

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Integrating [tex]\frac{1}{(1+sinx)}[/tex]

i just started learning Integration last week so not exactly sure how to approch this type of question.
 
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Do you know how to integrate 1/sinx?
 
no this is the first time i have seen where a trigonometric function is on the denominator

but now that i think about it

sinx = [tex]\sqrt{1-cos^2x}[/tex]
and arcsin was equal to [tex]\frac{1}{((1-x^2)}[/tex]

so i guess i could use the U substituition method
and then the answer would be...arcsin(cosx)? I am not too sure
 
Last edited:
No, no arcsin. But you might want to try multiplying numerator and denominator by (1-sin(x)). It may look more familiar.
 
Last edited:
You could also try the subsitution u = tan(x/2). It looks a little messy but everything cancels out.
 
well I am not sure if I am doing the right thing but here goes...

sinx = [tex]\frac{2tan(\frac{x}{2})}{1+tan^{2}(\frac{x}{2})}[/tex]

soo then i symplify the equation so that it is

[tex]\frac{1+tan^{2}(\frac{x}{2})}{1+tan^{2}(\frac{x}{2})+2tan(\frac{x}{2})}[/tex]

the i used u = tan[tex](\frac{x}{2})[/tex]

so i get [tex]\frac{1+u^{2}}{(u+1)^{2}}[/tex]

What should i do from here or is this the right way at all?
 
It looks to me like you are taking the long way around. Try multiplying numerator and denominator of your original problem by (1-sin(x)). You get (1-sinx)/cos^2(x). If you split that into two integrals you shouldn't have any problem with either of them.
 
Dick solution is quicker in this case but to integrate things like
1/(1+cosx+sinx) the substitution u=tan(x/2) is good.
But notice that it isn't x = tan(u/2) but rather u=tan(x/2). In order to get this into something the the example I gave you can show with so trig identities that if u=tan(x/2) then:
dx = 2du/(1+u^2)
sinx = 2u/(1+u^2)
cosx = (1-u^2)/(1+u^2)
If you substitue all that you the (1+u^2)s candel out and you get some rational function which you can solve with the typical rational function method (breaking it into elementry functions...)
 
wow by the way i got the answer using what dick said its actually pretty easy once u break it up...now I am going to try Daniels question going to see if i can get those now :) thnx a lot for the help by the way
the answe was tanx - secx +c
 

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